Bài tập về đa thức cho đơn thức.
1.
a) (21x2y−7xy2+xy):xy=21x−7y+1
b) (23x3y2−18x2y3+6xy5):4xy2=234x2−92xy+32y3
c) (5a3b2c+7a2b3c2−4abc2):a2bc=5ab+7b2c−4c
d) [6(x−2)3+7(x−2)4−9(x−2)5]:(2−x)2
= [6(x−2)3+7(x−2)4−9(x−2)5]:(x−2)2
= 6(x−2)+7(x−2)2−9(x−2)3
e) [5(x−y)6+2(x−y)5−7(x−y)3]:(y−x)2
= [5(x−y)6+2(x−y)5−7(x−y)3]:(x−y)2
= 5(x−y)4+2(x−y)3−7(x−y)
2. Đặt x2+5 = y ta có:
(x2+6)(x2+4):(x2+5)
= (y+1)(y−1):y
= (y2−1):y
= y−1y
⇒(x2+6)(x2+4):(x2+5)=x2+5−1x2+5
3.
a) [7(x2−1)4+2(1−x)3−3(x−1)2]:2(x−1)2
= [7(x−1)4(x+1)4−2(x−1)3−3(x−1)2]:2(x−1)2
= 72(x−1)2(x+1)4−(x−1)−32
b) [5(x3−y3)4+(x−y)3]:(x2−2xy+y2)
= [5(x−y)4(x2+xy+y2)4+(x−y)3]:(x−y)2
= 5(x−y)2(x2+xy+y2)4+(x−y)